Task: find all values of the parameter, such that integral converges. $$\int_0^{+\infty} \frac{dx}{1+x^a \sin^2x}$$ I tried a lot and I used Cauchy and Weierstrass method but it was useless. And now I know that I must use $$\sum_n \int_{\pi n}^{\pi(n+1)}$$


First, we have $$ \small\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+((k+1)\pi)^\alpha\sin^2(x)} \le\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)} \le\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+(k\pi)^\alpha\sin^2(x)}\tag{1} $$ Then to estimate the bounds in $(1)$, we have $$ \begin{align} \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} &=\int_0^\pi\frac{\mathrm{d}x}{1+\beta\sin^2(x)}\\ &=2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)}\tag{2} \end{align} $$ Noting that on $[0,\pi/2]$, $\frac2\pi x\le\sin(x)\le x$, we can bound $(2)$ $$ 2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta x^2} \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\frac{4\beta}{\pi^2}x^2}\tag{3} $$ Evaluating the bounds in $(3)$, we get $$ \frac2{\sqrt\beta}\arctan\left(\frac{\pi\sqrt\beta}2\right) \le2\int_0^{\pi/2}\frac{\mathrm{d}x}{1+\beta\sin^2(x)} \le\frac\pi{\sqrt\beta}\arctan\left(\sqrt\beta\right)\tag{4} $$ Combining $(1)$, $(2)$, and $(4)$, we get $$ \frac{\pi/2}{((k+1)\pi)^{\alpha/2}} \le2\int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)} \le\frac{\pi^2/2}{(k\pi)^{\alpha/2}}\tag{5} $$ Therefore, as $k\to\infty$, $$ \int_{k\pi}^{(k+1)\pi}\frac{\mathrm{d}x}{1+x^\alpha\sin^2(x)}=\Theta\left(k^{-\alpha/2}\right)\tag{6} $$ Thus, the integral converges for $\alpha\gt2$.


In the following, we shall use the following result twice: $\int \limits _0 ^\pi {1 \over {1 + c \sin ^2 x}} = {\pi \over \sqrt {1 + c}}$, if $c>0$. To prove it, split the integral into $\int \limits _0 ^{\pi \over 2} + \int \limits _{\pi \over 2} ^\pi$ and use the substitution $t = \tan x$ on each interval.

Note that $\int \limits _0 ^\infty {1 \over 1+x^a \sin ^2 x} \mathbb d x \geq \int \limits _0 ^\infty {1 \over 1+x^a} \mathbb d x$, and the latter integral is divergent for $a \leq 1$, therefore necessarily $a>1$.

Next, note that $$\int \limits _{n \pi} ^{(n+1) \pi} {1 \over 1+x^a \sin ^2 x} \mathbb d x = \int \limits _0 ^\pi {1 \over 1 + (x+n \pi)^a \sin ^2 x} \mathbb d x \leq \int \limits _0 ^\pi {1 \over 1+(n \pi)^a \sin ^2 x} \mathbb d x= {\pi \over \sqrt {1 + (n \pi)^a}}$$ and, since $\sum {\pi \over \sqrt {1 + (n \pi)^a}}$ converges if and only if $a>2$, your given integral also is guaranteed to converge for $a>2$.

It remains to investigate the case $1 < a \leq 2$. But $$\int \limits _0 ^\pi {1 \over 1 + (x+n \pi)^a \sin ^2 x} \mathbb d x \geq \int \limits _0 ^\pi {1 \over 1 + (\pi+n \pi)^a \sin ^2 x} \mathbb d x = {\pi \over \sqrt {1 + \pi ^a (n+1) ^a}}$$ and, since the series $\sum {\pi \over \sqrt {1 + \pi ^a (n+1) ^a}}$ is divergent for $a \leq 2$, so will be your integral.

Therefore, the series converges if and only if $a>2$.

  • $\begingroup$ Just curious, why using the $\mathbb{d}$ symbol for the $dx$? $\endgroup$ – Dmoreno May 29 '15 at 17:59
  • $\begingroup$ @Dmoreno: Agreed, I myself have given it some thought before making this choice: I want to stress that the letter "d" stands for something else than a variable or a parameter. For the same reason, if you want, for which we prefix "sin", "log", "exp" etc. by a "\". I use the same formatting for the numbers "e" and "i", to distinguish them from variables. $\endgroup$ – Alex M. May 29 '15 at 18:07
  • $\begingroup$ Oh, I see. That's the same reason for which I always try to type $\mathrm{d}x$ instead of $dx$. Cheers! $\endgroup$ – Dmoreno May 29 '15 at 18:11

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